To solve this problem, we need to find the largest possible integer $k$ such that for given sequences of side lengths $b_1 \leq b_2 \leq \ldots \leq b_{2009}$, $r_1 \leq r_2 \leq \ldots \leq r_{2009}$, and $w_1 \leq w_2 \leq \ldots \leq w_{2009}$, there are $k$ indices $j$ for which $b_j, r_j, w_j$ can form a non-degenerated triangle.

A non-degenerated triangle requires that the sum of the lengths of any two sides is greater than the length of the remaining side. This is known as the triangle inequality, and we need:

$$\begin{align*} b_j + r_j &> w_j, \\ r_j + w_j &> b_j, \\ w_j + b_j &> r_j. \end{align*}$$

Given that we sort each color's side lengths in non-decreasing order, the condition we must prove is that there exists at least one $j$ for which these inequalities hold true.

### Analysis

Consider a scenario where each set of lengths is from different triangles, the length $b_j, r_j, w_j$ may or may not satisfy the triangle inequalities due to their independent selection. However, since they are sorted sequences, permutations of indices alone without verification with the original triangles cannot guarantee a valid triangle due to the sorting operation mixing lengths from distinct triangles.

### Approach

To ensure that at least one combination of $(b_j, r_j, w_j)$ forms a valid triangle, let's consider the most restricted situation: the smallest non-trivial configuration is when $k = 1$, meaning we can ensure that a valid triangle can be formed with at least one combination due to sheer number and randomness in initial selections. Among $2009$ triangles, even under the worst permutations, there will always be at least one $j$ due to pigeonhole principle or inherent randomness that keeps at least one triple in a potentially valid configuration for the triangle inequalities.

### Conclusion

Thus, the largest integer $k$ for which we can guarantee these side combinations form at least one valid non-degenerated triangle is:

$$\boxed{1}$$